F.3 MATH ( 三角學的應用 )

Q1
斜率=PQ的高度變化/PQ的平移距離=(250-225)/300=(250-225):300 and so on

Q2
(a)theta = angle ACB = tan^-1 (AB/CB)
全方位角=360degree - theta
象限角=N theta W
(b)畢氏定理=> AC = 開方(AB^2 + CB^2)

Q3
(a)theta = AB的斜坡傾角=tan^-1 (AB的斜率)
sin theta = AB的上升距離/AB的斜面距離
=> AB的上升距離 = sin theta * AB的斜面距離
(b)
BC的上升距離 = AC的上升距離 - AB的上升距離
BC的斜坡傾角 = phi = sin^-1 (BC的上升距離/BC的斜面距離)
BC的斜率 = tan phi

Q4
AB兩廈的高度差距 = AB兩廈之間的距離 * tan 27.3degree
A廈高度 = B廈高度 - AB兩廈的高度差距
由B廈底對A廈的仰角 = tan^-1 (A廈高度 / AB兩廈之間的距離)

2010-04-11 19:28:06 補充：
如你要求上面的是式

2010-04-11 19:29:41 補充：
人家叫你畀式咋...你畀埋答案, 咁人家唔駛做la...

1. PQ 上升距離=250-225=25m
PQ 斜率 = 25/300 = 1:12

2a tan a = 350/470 a = 36.67度b= 90 - 36.67 = 53.3度
由A測得C的方位是北53.3度西
b AC=(square root) (350^2+470^2) = 586 km

3a BD/50=0.15 BD=7.5m
b BE= (sq root) (40^2-16^2) = 36.66 m
BC 斜率 = 16/BE = 0.44

4 tan 27.3度 = x/20 x=20tan27.3度
y = 40 - x = 40 - 20tan27.3度
tan a = y/20
a = 56.0 度
B測得A的頂部的仰角是56.0度.
圖片參考：http://imgcld.yimg.com/8/n/HA00123784/o/701004100139413873382280.jpg


2010-04-11 19:22:17 補充：
I have a quiz about trigo tomorrow, so I just treat it as a revision. I think next time you should try to do the questions first, as they are not really difficult.

5 marks for such a long question ????? Who would be so stupid to answer you???